CODEWITHSUNDEEP
pythonpandas
Charizard_knows_to_code
Charizard_knows_to_code

Reputation: 430

Add timedelta to date column based on conditions

Here is my sample data

        Date    tf  tf_int
0   2022-01-01  D   1
1   2022-01-02  W   3
2   2022-01-03  M   2
3   2022-01-04  Y   2

I want to make a new column based that will add the required timedelta to the Date column.

For eg, for my first row, I want to add 1 day to the date (2022-1-1) so the result would be 2022-1-2

Is there any vectorised way to achieve this ? I don't using multiple .loc[] calls to accomplish this separately for days,weeks, months etc; But I would prefer the solution to be vectorised

My unsuccessful attempt with .loc[]

df.loc[df['tf'] =='D','publish_date'] = df['Date'].dt.date + dt.timedelta(days = df['tf_int'])

But this results in error because df['tf_int'] is a series and not an exact value

Edit: For my use case, adding one month will add affect only the month component, and not do anything to the date

Upvotes: 1

Views: 132

Answers (2)

mozway
mozway

Reputation: 262254

If you accept to have approximations for Month/Year, a vectorial solution could be something like:

mapper = {'D': 1, 'W': 7, 'M': 30, 'Y': 365}
df['Date2'] = (pd.to_datetime(df['Date'])
                 .add(pd.to_timedelta(df['tf'].map(mapper).mul(df['tf_int']),
                                      unit='D'))
              )

output:

        Date tf  tf_int      Date2
0 2022-01-01  D       1 2022-01-02
1 2022-01-02  W       3 2022-01-23
2 2022-01-03  M       2 2022-03-04
3 2022-01-04  Y       2 2024-01-04

Upvotes: 0

jezrael
jezrael

Reputation: 863501

If need exact year and months is necessary use list comprhension solution with offsets.DateOffset:

d = {'D':'days','W':'weeks', 'M':'months', 'Y':'years'}

df['publish_date'] = [z + pd.offsets.DateOffset(**{d[y]: x}) 
                     for x, y, z in zip(df['tf_int'], df['tf'], df['Date'].dt.normalize())]
print (df)
        Date tf  tf_int publish_date
0 2022-01-01  D       1   2022-01-02
1 2022-01-02  W       3   2022-01-23
2 2022-01-03  M       2   2022-03-03
3 2022-01-04  Y       2   2024-01-04

Upvotes: 2

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